A particular parallel-plate capacitor has air (K-1) between the plates. The capa

Question

A particular parallel-plate capacitor has air (K-1) between the plates. The capacitor is charged by connecting it to a 12 V battery, and then the battery is disconnected from the capacitor. At this instant in time, the electric field between the plates is Eo, and the energy stored is Uo A slab of dielectric material, with a dielectric constant of 4, in now inserted into the capacitor, completely filling the space between the capacitor plates. [2 points] (a) In terms of Eo, the electric field between the plates is now Show your work/thinking here) 2 points] (b) The voltage across the plates is now Show your work/thinking here) 2 points] (c) In terms of Uo, the energy stored in the capacitor is now Show your work/thinking here) 2 points] (d) With the slab of dielectric still between the plates, the capacitor is then re-connected to the 12-volt battery. In terms of UD, the energy stored in the capacitor is now Show your work/thinking here)

Explanation / Answer

a)

C = capacitance

d = distance between the plates

V = voltage across the capacitor

charge stored in capacitor

Q = CV

electric field between the plates is given as

Eo = V/d

energy stored is given as

Uo = (0.5) C V2

when dielectric is inserted

C' = new capacitance = 4C

V' = new capcitance between the plates

Since charge in capacitor remain same

Q' = Q

C' V' = CV

4C V' = CV

V' = V/4

electric field between the plates is given as

E'o = V' /d = (V/4 )/d = Eo /4

b)

V' = V/4 = 12 /4 = 3 Volts

c)

U' = new energy stored = (0.5) C'V'2 = (0.5) 4 C (V/4)2 = Uo/16

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